1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)

\(x+\frac{4}{5}=\pm\frac{4}{5}\)

\(TH1:x+\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)

\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)

\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)

Vậy x ∈ {0; \(\frac{-8}{5}\)}

Hai câu cuối khó nhìn nên ko giải

a,

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)

\(\Rightarrow5^{x+3}2=2.5^{11}\)

\(\Rightarrow5^{x+3}=5^{11}\)

\(\Rightarrow x+3=11\)

\(\Rightarrow x=8\)

b, (Check lai xem de sai o dau khong nhe)

\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)

Dat 5x ra ben ngoai

\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)

\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)

\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)

\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)

\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)

\(\Rightarrow5^x\left(5^{-3}\right).9379\)

=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)

B(3)=2*3^2-4*3+3=18-12+3=9

B(-1/2)=2*1/4-4*(-1/2)+3=1/2+3+2=1/2+5=11/2

e tách ra nhé.☘

;))

a: 2x-3/2+3/4=-4

=>2x-3/4=-4

=>2x=-13/4

hay x=-13/8

b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)

=>2/3x+3/5=12/145

=>2/3x=-15/29

hay x=-45/58

c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)

=>1/2x-3/5x+13/5=-7/10x-7/5

=>-1/10x+7/10x=-7/5-13/5

=>3/5x=-2

hay x=-2:3/5=-10/3

A = 2⁵.(-5)² - 8² - 7

= 32.25 - 64 - 7

= 729

= 27²

B = 2³.(-4)² + (-3)².3² - 40

= 8.16 + 9.9 - 40

= 169

= 13²

C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³

= (-5/4)³ . (8/5)³

= (-5/4 . 8/5)³

= (-2)³

D = (-1/4)² : (1/2 - 1/3)

= 1/16 : 1/6

= 3/8

E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³

= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³

= 1/4 + 25 . (15/16)³ : 27/8

= 1/4 + 25 . 3375/4096 : 27/8

= 1/4 + 84375/4096 : 27/8

= 1/4 + 3125/512

= 3253/512

F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8

= 8 + 3.1 - 1 + (4 : 1/2) - 8

= 8 + 3 - 1 + 8 - 8

= 10

Nguyễn Trà My

Phần a)

\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(32-3x+13=76-x\)

\(116-3x=76-x\)

\(116-76=3x-x\)

\(46=2x\)

\(x=46\div2\)

\(x=13\)

a)  \(3.\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(3.\left(\frac{1}{2}-x\right)+x=\frac{7}{6}-\frac{1}{3}\)

\(\Rightarrow\frac{3}{2}-3x+x=\frac{5}{6}\)

\(-3x+x=\frac{5}{6}-\frac{3}{2}\)

\(2x=-\frac{2}{3}\)

\(x=-\frac{2}{3}:2\)

\(x=-\frac{1}{3}\)

\(=\dfrac{\left(\dfrac{2}{3}\cdot\dfrac{-3}{4}\right)^2\cdot\dfrac{2}{3}\cdot\left(-1\right)}{\left(\dfrac{2}{5}\cdot\dfrac{-5}{12}\right)^2}=\dfrac{\dfrac{1}{4}\cdot\dfrac{2}{3}\cdot\left(-1\right)}{\dfrac{1}{36}}=\dfrac{-2}{12}:\dfrac{1}{36}=-\dfrac{1}{6}\cdot36=-6\)

thank iu bạn nha

:((

1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)

\(\Leftrightarrow-2x^2-2x=0\)

\(\Leftrightarrow-2x\left(x+1\right)=0\)

Vì -2≠0

nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy: x∈{0;-1}

2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)

\(\Leftrightarrow11x-2=0\)

\(\Leftrightarrow11x=2\)

hay \(x=\frac{2}{11}\)

Vậy: \(x=\frac{2}{11}\)

3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)

\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)

\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)

hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)

Vậy: \(x=\frac{-13}{4}\)

4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)

1. \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x=3x^2-x\)

\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)

\(\Leftrightarrow-2x-2x^2=0\)

\(\Leftrightarrow-2x\left(1+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

2. \(15-5\left(1-2x\right)=12-x\)

\(\Leftrightarrow15-5+10x=12-x\)

\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)

\(\Leftrightarrow-2+11x=0\)

\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)

3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)

\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)

\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)

4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)

\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)

\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)